# Solving equations with powers without logarithms

Im taking an introduction to logarithms. Of course a short review of exponentiation is inherent for a clear understanding of logarithms.

I was asked to find, for example, $27^x = 3$. (without the use of logarithms) Am i to use pure intuition that 3 is a factor of 27?

Lastly, I was asked to find $2^{2x} - 6 * 2^{x} + 8 = 0$. It is obvious this is a quadratic equation, however how do i particularly use the formula to solve for the x power.

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For the first one, you can observe that 3 is a factor of 27, or you can just take logs of both sides and simplify. –  vadim123 Jul 12 '14 at 17:16
For the first one, isn't the natural thing to write just $\sqrt[3]{27}=3$? –  MPW Jul 12 '14 at 17:40

For your first example, you can either use logarithms as in $$x \log 27 = \log 3$$ or your knowledge that $27=3^3$ so $$3^{3x}=3^1 \text { so } 3x=1$$ or some combination of the two so $$3x \log 3= \log 3.$$
For the second, $2^{2x}=\left(2^x\right)^2$ so you can turn this into a quadratic you can solve.
For the second equation, take $y=2^x$. Then your equation becomes $$y^2-6y+8=0$$